Proof. Since $\dim \mathcal {X}^\bot _-=1$ we know that $\mathcal {X}_-\subsetneqq \mathcal {X}$ . Condition 1 implies that $\mathcal {X}^\bot _-=\mathcal {X}^\bot _-(\mathcal {A})$ . For any $y\in \mathcal {X}\setminus \mathcal {X}_-$ , by the Hahn–Banach theorem (pick $\widetilde f\in (\mathcal {X}/\mathcal {X}_-)^*$ such that $\widetilde f(y+\mathcal {X}_-)=1$ , and let $f=\widetilde f\circ \theta $ , where $\theta $ is the quotient mapping), it follows that there exists $f\in \mathcal {X}^\bot _-$ such that $f(y)=1$ . Since $\dim \mathcal {X}^\bot _-(\mathcal {A})=1$ and $\mathcal {X}(\mathcal {A})=\mathcal {X}$ , we get $x\otimes f\in \mathcal {A} \ (\text {for all } x\in \mathcal {X})$ . Thus, for any $A\in \mathcal {Z_A}(\mathcal {B(X)})$ , we have $A(x\otimes f)y=(x\otimes f)Ay$ . That is, $Ax=f(Ay)x$ . Because x is arbitrary, we conclude that $A\in \mathbb {C}I$ .

Proof. By the assumption $\dim \mathcal {X}^\bot _-=1$ and Condition 1, it follows that $\mathcal {X}^\bot _-=\mathcal {X}^\bot _-(\mathcal {A})$ . If $\text {codim}\, \mathcal {X}_->1$ , then there are linearly independent elements $x_1+\mathcal {X}_-$ and $x_2+\mathcal {X}_-$ in $\mathcal {X}/\mathcal {X}_-$ . By a proof analogous to that of Lemma 2.1, one can construct $f_1,f_2\in \mathcal {X}^\bot _-$ satisfying $f_i(x_j)=\delta _{ij}$ , where $\delta _{ij}$ is the Kronecker delta. Then $f_1, f_2$ are linearly independent, which contradicts that $\dim \mathcal {X}^\bot _-=1$ . Hence, ${\text {codim}\, \mathcal {X}_-=1}$ . Based on this fact, $\ker f=\mathcal {X}_-$ for any $f\in \mathcal {X}^\bot _-\setminus \{0\}$ . Fixing a certain $f_0\in \mathcal {X}^\bot _-(\mathcal {A})\setminus \{0\}=\mathcal {X}^\bot _-\setminus \{0\}$ , by Condition 2 there exists $y\in \mathcal {X}\setminus \mathcal {X}^-$ such that $y\otimes f_0\in \mathcal {A}$ . Denote $M=\text {span}\,\{y\}$ , and then $\mathcal {X}=\mathcal {X}_-\oplus M$ . Indeed, since $\sigma \colon M\longrightarrow \mathcal {X}/\mathcal {X}_-$ , $y\longmapsto y+\mathcal {X}_-$ is an isomorphism and $\sigma _f\colon \mathcal {X}/\mathcal {X}_-\longrightarrow \mathbb {C}$ , $y+\mathcal {X}_-\longmapsto f(y)$ is also an isomorphism for any $f\in \mathcal {X}^\bot _-\setminus \{0\}$ (via the fundamental theorem of homomorphisms), so we have $\sigma_f \circ \sigma: M \to \mathbb{C}$ is an isomorphism. Let $x\in \mathcal {X}$ and $f\in \mathcal {X}^\bot _-\setminus \{0\}$ . Then there exists $y\in M$ such that $f(y)=\sigma _f\circ \sigma (y)=f(x)$ . Hence $x-y\in \ker f=\mathcal {X}_-$ ; whereupon $\mathcal {X}\subseteq \mathcal {X}_-+M$ .

Choose $x_0\in M$ such that $f_0(x_0)=1$ , define $P=x_0\otimes f_0\in \mathcal {A}$ and put $Q=I-P$ . Then P and Q are idempotents satisfying $PQ=QP=0$ , $P\mathcal {X}=M$ , and $Q\mathcal {X}=\mathcal {X}_-$ . We get the Peirce decomposition

$$ \begin{align*} \mathcal {B(X)}=P\mathcal {B(X)}P\oplus P\mathcal {B(X)}Q\oplus Q\mathcal {B(X)}P\oplus Q\mathcal {B(X)}Q. \end{align*} $$

For any $A\in \mathcal {A}$ , since $\dim \mathcal {X}^\bot _-=1$ and $A^*f_0\in \mathcal {X}^\bot _-=\mathcal {X}^\bot _-(\mathcal {A})$ , there exists $\unicode{x3bb} _A\in \mathbb {C}$ such that $A^*f_0=\unicode{x3bb} _Af_0$ . Then

Thus $\mathcal {A}=\mathcal {A}_{11}\oplus \mathcal {A}_{21}\oplus \mathcal {A}_{22}$ , where $\mathcal {A}_{11}=P\mathcal {A} P$ , $\mathcal {A}_{21}=Q\mathcal {A} P$ , $\mathcal {A}_{22}=Q\mathcal {A} Q$ .

We divide the proof of the assertion into two cases.

Case 1: n is odd. For an arbitrary $A\in \mathcal {A}_{21}$ , we have $AQ=0$ and $QA=A$ . A direct computation shows that

(3-1) $$ \begin{align} L(A)&=L(p_n(A,Q,\ldots,Q))\nonumber\\[-6pt] &=p_n(L(A),Q,\ldots,Q)+\sum_{i=2}^{n-1}p_n(A,\ldots,Q,\mathop{L(Q)} _{\substack{\uparrow\\[-3pt] i\,\text{th}}},Q, \ldots,Q)\nonumber\\[-3pt] &\quad+p_n(A,\underbrace{Q,\ldots,Q}_{n-2}, L(Q))\nonumber\\[-3pt] &=QL(A)P+PL(A)Q-(n-2)(AL(Q)-AL(Q)Q-QL(Q)A)\nonumber\\[-3pt] &\quad-AL(Q)+L(Q)A. \end{align} $$

Premultiply (respectively, postmultiply) the relation (3-1) by Q (respectively P) to get

$$ \begin{align*} QL(Q)A=AL(Q)P. \end{align*} $$

For any $x\in \mathcal {X}_-$ , since $\mathcal {X}_-\subseteq \mathcal {X}(\mathcal {A})$ and $\mathcal {X}^\bot _-(\mathcal {A})=\mathcal {X}^\bot _-$ is one-dimensional, we see that $A_0=x\otimes f_0\in \mathcal {A}$ . Obviously, $A_0=QA_0P\in \mathcal {A}_{21}$ . So

(3-2) $$ \begin{align} QL(Q)A_0=A_0L(Q)P=(x\otimes f_0)L(Q)(x_0\otimes f_0)=f_0(L(Q)x_0)(x\otimes f_0). \end{align} $$

Taking into account the action of (3-2) on $x_0$ , we have

$$ \begin{align*}QL(Q)x=f_0(L(Q)x_0)x. \end{align*} $$

Since $x\in \mathcal {X}_-$ is arbitrary,

(3-3) $$ \begin{align} QL(Q)Q=f_0(L(Q)x_0)Q. \end{align} $$

A direct verification gives

(3-4) $$ \begin{align} PL(Q)P=(x_0\otimes f_0)L(Q)P=f_0(L(Q)x_0)P. \end{align} $$

Combining (3-3) with (3-4), we obtain

$$ \begin{align*}QL(Q)Q+PL(Q)P=f_0(L(Q)x_0)I. \end{align*} $$

Write $T=PL(Q)Q-QL(Q)P$ , and define $D_T(\diamond )=[T,\diamond ]$ . Then $D_T$ is a (continuous) derivation of $\mathcal {A}$ satisfying the condition

$$ \begin{align*} L(Q)=D_T(Q)+f_0(L(Q)x_0)I. \end{align*} $$

Put $\widetilde L=L-D_T$ . Then $\widetilde L$ is a Lie n-derivation on $\mathcal {A}$ and $\widetilde L(Q)=f_0(L(Q)x_0)I$ . A straightforward calculation yields that

$$ \begin{align*} \widetilde L(A)&=\widetilde L(p_n(A,Q,\ldots,Q))\\ &=p_n(\widetilde L(A),Q,\ldots, Q)+p_n(A,Q,\ldots,Q,\widetilde L(Q))\\ &=Q\widetilde L(A)P+P\widetilde L(A)Q. \end{align*} $$

For any $A'\in \mathcal {A}_{21}$ and $B\in \mathcal {A}$ ,

$$ \begin{align*} 0&=\widetilde L(p_n(A,A',B,\ldots,B))\\ &=p_n(\widetilde L(A),A',B,\ldots,B)+p_n(A,\widetilde L(A'),B,\ldots,B)\\ &=p_{n-1}([\widetilde L(A),A']+[A,\widetilde L(A')], B,\ldots,B). \end{align*} $$

By Lemma 2.6, Lemma 3.2 and a recursive computation, we eventually arrive at $[\widetilde L(A), A']+[A,\widetilde L(A')]=\unicode{x3bb} I$ , where $\unicode{x3bb} \in \mathbb {C}$ . Thus

$$ \begin{align*} [Q\widetilde L(A)P+P\widetilde L(A)Q,A']&=[\widetilde L(A),A']\\ &=\unicode{x3bb} I-[A,\widetilde L(A')]\\ &=\unicode{x3bb} I-p_n(-A,Q,\ldots, Q,\widetilde L(A'))\\ &=\unicode{x3bb} I-\widetilde L(p_n(-A,Q,\ldots,Q,A'))+p_n(\widetilde L(-A),Q,\ldots,Q,A')\\ &=\unicode{x3bb} I-\widetilde L(p_2(A,A'))+p_3(Q,\widetilde L(A),A')\\ &=\unicode{x3bb} I+p_3(Q,\widetilde L(A),A')\\ &=\unicode{x3bb} I+p_3(Q,Q\widetilde L(A)P+P\widetilde L(A)Q,A')\\ &=\unicode{x3bb} I+[Q\widetilde L(A)P-P\widetilde L(A)Q,A']. \end{align*} $$

We therefore have $[P\widetilde L(A)Q, A']\kern1pt{=}\kern1pt\tfrac 12\unicode{x3bb} I$ . By Lemma 2.6, it follows that ${[P\widetilde L(A)Q, A']\kern1pt{=}\kern1pt 0}$ . Premultiply it by P to get $P\widetilde L(A)QA'=0$ . In particular, $P\widetilde L(A)QA_0=0$ . Applying the equality to $x_0$ , we obtain $P\widetilde L(A)Qx=0$ . Note that $QM=\{0\}$ , and thus $P\widetilde L(A)Q=0$ . Hence $\widetilde L(A)=Q\widetilde L(A)P\in \mathcal {B}_{21}$ .

For an arbitrary $A\in \mathcal {A}_{22}$ ,

$$ \begin{align*} 0=\widetilde L(p_n(A,Q,\ldots,Q))=p_n(\widetilde L(A),Q,\ldots,Q)=Q\widetilde L(A)P+P\widetilde L(A)Q. \end{align*} $$

So $0=Q\widetilde L(A)P=P\widetilde L(A)Q$ . This gives that $\widetilde L(A)=Q\widetilde L(A)Q+P\widetilde L(A)P$ . For any $B\in \mathcal {A}_{11}$ and $A'\in \mathcal {A}$ , we get

$$ \begin{align*} 0=\widetilde L(p_n(A,B,A',\ldots,A'))=p_{n-1}([\widetilde L(A),B]+[A,\widetilde L(B)],A',\ldots,A'). \end{align*} $$

Applying Lemma 2.6 and Lemma 3.2 and using a recursive computation, we conclude that $[\widetilde L(A),B]+[A,\widetilde L(B)]=\mu I$ , where $\mu \in \mathbb {C}$ . Since $B\in \mathcal {A}_{11}$ , we see that $[\widetilde L(A),B]=[P\widetilde L(A)P,B]$ . Thus we obtain

$$ \begin{align*} [P\widetilde L(A)P,B]=P([P\widetilde L(A)P,B]+[A,\widetilde L(B)])P=\mu P. \end{align*} $$

Looking at the above equation as an operator equation over $P\mathcal {X}=M$ and exploiting Lemma 2.6, we see that $\mu =0$ . In particular, we take $B=x\otimes f_0\in \mathcal {A}_{11}$ , where $x\in M$ . Considering the action of $[P\widetilde L(A)P,x\otimes f_0]=0$ on $x_0$ ,

$$ \begin{align*} P\widetilde L(A)P=H_{22}(A)P, \end{align*} $$

where $H_{22}(A)=f_0(P\widetilde L(A)x_0)I\in \mathbb {C}I$ . The preceding equality shows the linearity and boundedness (if L is bounded) of $H_{22}(A)$ for all $A\in \mathcal {A}_{22}$ . Moreover, we have

$$ \begin{align*} \widetilde L(A)-H_{22}(A)&=Q\widetilde L(A)Q+P\widetilde L(A)P-H_{22}(A)(P+Q)\\ &=Q\widetilde L(A)Q-H_{22}(A)Q\in\mathcal{B}_{22}. \end{align*} $$

In an analogous manner, one can prove that, for any $A\in \mathcal {A}_{11}$ , there exists a (bounded if L is bounded) linear operator $H_{11}\colon \mathcal {A}_{11}\to \mathbb {C}I$ such that $\widetilde L(A)-H_{11}(A)\in \mathcal {B}_{11}$ .

Case 2: n is even. For an arbitrary $A\in \mathcal {A}_{21}$ , a direct calculation shows that

(3-5) $$ \begin{align} -L(A)&=L(p_n(A,Q,\ldots,Q))\nonumber\\ &=p_n(L(A),Q,\ldots,Q)+\sum_{i=2}^{n-1}p_n(A,\ldots,Q,\mathop{L(Q)} _{\substack{\uparrow\\ i\,\text{th}}},Q, \ldots,Q)\nonumber\\ &\quad+p_n(A,\underbrace{Q,\ldots,Q}_{n-2}, L(Q))\nonumber\\ &=p_2(L(A),Q)-(n-2)p_3(A,L(Q),Q)+[A,L(Q)]\nonumber\\ &=L(A)Q-QL(A)-(n-2)(AL(Q)Q-AL(Q)+QL(Q)A)\nonumber\\ &\quad+AL(Q)-L(Q)A. \end{align} $$

Premultiply (3.1) by Q and postmultiply (3.1) by P to obtain

$$ \begin{align*} QL(Q)A=AL(Q)P. \end{align*} $$

Adopting the same discussion as in Case 1 gives

$$ \begin{align*} L(Q)=[T,Q]=f_0(L(Q)x_0)I. \end{align*} $$

Let us write $D_T(\diamond )=[T, \diamond ]$ and $\widetilde L=L-D_T$ . Then $\widetilde L(Q)=f_0(L(Q)x_0)I$ . Thus

$$ \begin{align*} -\widetilde L(A)&=\widetilde L(p_n(A,Q\ldots,Q))\\ &=p_n(\widetilde L(A),Q,\ldots,Q)\\ &=p_2(\widetilde L(A),Q)\\ &=\widetilde L(A)Q-Q\widetilde L(A)\\ &=P\widetilde L(A)Q-Q\widetilde L(A)P. \end{align*} $$

So $\widetilde L(A)=Q\widetilde L(A)P-P\widetilde L(A)Q$ . For any $A'\in \mathcal {A}_{21}$ and $B\in \mathcal {A}$ , a straightforward computation results in

$$ \begin{align*} 0&=\widetilde L(p_n(A,A',B,\ldots,B))\\ &=p_n(\widetilde L(A),A',B,\ldots,B)+p_n(A,\widetilde L(A'),B,\ldots,B)\\ &=p_{n-1}([\widetilde L(A),A']+[A,\widetilde L(A')],B,\ldots,B). \end{align*} $$

By Lemma 2.6, Lemma 3.2 together with a recursive computation, we obtain

(3-6) $$ \begin{align} [\widetilde L(A),A']+[A,\widetilde L(A')]=\unicode{x3bb} I, \end{align} $$

where $\unicode{x3bb} \in \mathbb {C}$ . Hence

$$ \begin{align*} [\widetilde L(A),A']&=[Q\widetilde L(A)P-P\widetilde L(A)Q,A']\\ &=\unicode{x3bb} I-[A,\widetilde L(A')]\\ &=\unicode{x3bb} I-[p_{n-1}(A,Q,\ldots,Q),\widetilde L(A')]\\ &=\unicode{x3bb} I-\widetilde L(p_n(A,Q,\ldots,Q,A'))+p_n(\widetilde L(A),Q,\ldots,Q,A')\\ &=\unicode{x3bb} I-p_2(A,A')+[p_3(\widetilde L(A),Q,Q),A']\\ &=\unicode{x3bb} I+[-P\widetilde L(A)Q+Q\widetilde L(A)P,A']. \end{align*} $$

This forces $\unicode{x3bb} =0$ . It follows from the relation (3-6) that

(3-7) $$ \begin{align} -P\widetilde L(A)A'+A'\widetilde L(A)Q+A\widetilde L(A')-\widetilde L(A')A=0. \end{align} $$

Postmultiply the relation (3-7) by Q to produce $(\widetilde L(A)A'+\widetilde L(A')A)Q=0$ . Premultiply the relation (3-7) by P obtain $P(\widetilde L(A)A'+\widetilde L(A')A)=0$ . Put $S=\widetilde L(A)A'+\widetilde L(A')A$ . Then

$$ \begin{align*} SQ=0,\;PS=0. \end{align*} $$

On the other hand, since $\widetilde L(A')=Q\widetilde L(A')P-P\widetilde L(A')Q$ , by (3-6) we see that

$$ \begin{align*} -P\widetilde L(A)A'+A'\widetilde L(A)Q-A\widetilde L(A')Q+P\widetilde L(A')A=0. \end{align*} $$

Thus

$$ \begin{align*} \mathcal{B}_{11}\ni P(-\widetilde L(A)A'+\widetilde L(A')A)P=Q(A\widetilde L(A')-A'\widetilde L(A))Q\in\mathcal{B}_{22}. \end{align*} $$

Therefore, $P\widetilde L(A)A'=P\widetilde L(A')A$ and $A\widetilde L(A')Q=A'\widetilde L(A)Q$ . Applying the fact $PS=0$ yields that $P\widetilde L(A)A'=0$ . In particular, for $A'=x\otimes f_0\in \mathcal {A}_{21}$ , where $x\in \mathcal {X}_-$ , the relation

$$ \begin{align*} 0=P\widetilde L(A)A'=P\widetilde L(A)Q(x\otimes f_0) \end{align*} $$

holds true. Considering its action on $x_0$ , we observe that $P\widetilde L(A)Qx=0$ . Since $QM=0$ , we get $P\widetilde L(A)Q=0$ and hence $\widetilde L(A)=Q\widetilde L(A)P\in \mathcal {B}_{21}$ .

For an arbitrary $A\in \mathcal {A}_{22}$ , a direct calculation gives rise to

$$ \begin{align*}\begin{aligned} 0&=\widetilde L(p_n(A,Q,\ldots,Q))\\ &=p_n(\widetilde L(A),Q,\ldots,Q)\\ &=p_2(\widetilde L(A),Q)\\ &=P\widetilde L(A)Q-Q\widetilde L(A)P. \end{aligned} \end{align*} $$

This implies that $Q\widetilde L(A)P=P\widetilde L(A)Q=0$ . Thus $\widetilde L(A)=Q\widetilde L(A)Q+P\widetilde L(A)P$ . The remainder of the deduction and discussion for $A\in \mathcal {A}_{11}$ are identical with the counterparts of Case 1, and so we do not repeat it here. This completes the proof of the assertion.

We now return to the proof of Theorem 3.3. For an arbitrary $A=A_1+A_2+A_3\in \mathcal {A}=\mathcal {A}_{11}\oplus \mathcal {A}_{21}\oplus \mathcal {A}_{22}$ , we establish a mapping

$$ \begin{align*}\begin{aligned} H\colon\mathcal{A}&\longrightarrow \mathbb{C}I\\ A&\longmapsto (H_{11}(A_1)+H_{22}(A_3))I. \end{aligned} \end{align*} $$

The linearity of $H_{ii}\,(i=1,2)$ implies that H is linear. If L is continuous, by the properties of direct products, it is not difficult to verify the continuity of H. Define a mapping $\widetilde D\colon \mathcal {A}\to \mathcal {B(X)}$ via the relation $\widetilde D=\widetilde L-H$ . Then $\widetilde D$ is a linear mapping such that $\widetilde D(A_1)\in \mathcal {B}_{11}$ , $\widetilde D(A_2)\in \mathcal {B}_{21}$ , and $\widetilde D(A_3)\in \mathcal {B}_{22}$ .

We next demonstrate that $\widetilde D$ is a derivation. Since

$$ \begin{align*} \mathcal{A}_{21}\ni A_3A_2=[A_3,A_2]=[[A_2,Q],A_3]=(-1)^{n-1}p_n(A_2,Q,\ldots,Q,A_3), \end{align*} $$

we see that

$$ \begin{align*} \widetilde D(A_3A_2)&=\widetilde L(A_3A_2)\\ &=(-1)^{n-1}(p_n(\widetilde L(A_2),Q,\ldots,Q,A_3)+p_n(A_2,Q,\ldots,Q,\widetilde L(A_3)))\\ &=p_2(A_3,\widetilde L(A_2))+p_2(\widetilde L(A_3),A_2)\\ &=p_2(A_3,\widetilde D(A_2))+p_2(\widetilde D(A_3),A_2)\\ &=A_3\widetilde D(A_2)+\widetilde D(A_3)A_2. \end{align*} $$

For an arbitrary $A_3'\in \mathcal {A}_{22}$ , it follows that

(3-8) $$ \begin{align} \widetilde D(A_3A_3'A_2)=\widetilde D((A_3A_3')A_2)=\widetilde D(A_3A_3')A_2+A_3A_3'\widetilde D(A_2). \end{align} $$

On the other hand,

(3-9) $$ \begin{align} \widetilde D(A_3A_3'A_2)=\widetilde D(A_3(A_3'A_2))=\widetilde D(A_3)A_3'A_2+A_3\widetilde D(A_3')A_2+A_3A_3'\widetilde D(A_2). \end{align} $$

Comparing (3-8) with (3-9) yields

(3-10) $$ \begin{align} \widetilde D(A_3A_3')A_2=(\widetilde D(A_3)A_3'+A_3\widetilde D(A_3'))A_2. \end{align} $$

In particular, we take $A_2=x\otimes f_0$ , where $x\in \mathcal {X}_-$ . Using (3-10) to act on $x_0$ , we obtain

$$ \begin{align*} \widetilde D(A_3A_3')x=(\widetilde D(A_3)A_3'+A_3\widetilde D(A_3'))x. \end{align*} $$

In view of the fact $QM=\{0\}$ , we know that

$$ \begin{align*} \widetilde D(A_3A_3')=\widetilde D(A_3)A_3'+A_3\widetilde D(A_3'). \end{align*} $$

For the treatment of

$$ \begin{align*} \mathcal{A}_{21}\ni A_2A_1=[A_2,A_1]=(-1)^np_n(A_2,Q,\ldots,Q,A_1), \end{align*} $$

the discussion is parallel to that for $A_3A_2$ above, and for $A_1A_1'$ (where $A_1'\in \mathcal {A}_{11}$ ), the process is totally analogous to the above discussion of $A_3A_3'$ . We therefore omit them here.

For any elements $A=A_1+A_2+A_3$ , $B=B_1+B_2+B_3\in \mathcal {A}=\mathcal {A}_{11}\oplus \mathcal {A}_{21}\oplus \mathcal {A}_{22}$ , we eventually get

$$ \begin{align*}\begin{aligned} \widetilde D(AB)&=\widetilde D(A_1B_1+A_2B_1+A_3B_2+A_3B_3)\\ &=\widetilde D(A)B+A\widetilde D(B). \end{aligned} \end{align*} $$

This shows that $\widetilde D$ is a derivation. Let $D=\widetilde D+D_T$ (where $D_T$ is the derivation established in the previous assertion). Then D is a (continuous whenever L is continuous) derivation, and L has the following decomposition:

$$ \begin{align*} L=\widetilde L+D_T=\widetilde D+H+D_T=D+H. \end{align*} $$

By Proposition 2.5, we conclude that $H\colon \mathcal {A}\longrightarrow \mathbb {C}I$ vanishes on each $(n-1)\,$ th commutator on $\mathcal {A}$ .

Proof. The case of $f=0$ is trivial. Suppose that $f\in \mathcal {X}^\bot _-(\mathcal {A})\setminus \{0\}$ . If $x\in \ker f$ , by Condition 2, there exists $y\in \mathcal {X}\setminus \mathcal {X}_-$ such that $f(y)=1$ and $y\otimes f\in \mathcal {A}$ . Since $\dim \mathcal X>2$ , one has $\dim \ker f>1$ and thus can find $x'\in \ker f$ linearly independent from x. Put $y'=y+x'$ . Then $y'\in \mathcal {X}(\mathcal {A})$ and $y, y', x$ are linearly independent. Indeed, suppose that $ax+by+cy'=0$ for some $a,b,c\in \mathbb C$ . Applying f to this equation yields $b+c=0$ . We therefore have

$$ \begin{align*} 0=ax+by+cy'=ax+(b+c)y+cx'=ax+cx'. \end{align*} $$

The linear independence of $x,x '$ gives that $a=c=0$ . Hence $b=0$ as well. For any $z\in \ker f$ , a direct computation shows that

$$ \begin{align*} L(x\,\otimes\, f)z&=L(p_n(x\otimes f,y\otimes f,\ldots,y\otimes f))z\\ &=p_n(L(x\otimes f),y\otimes f,\ldots,y\otimes f)z+p_n(x\otimes f,y\otimes f,\ldots,y\otimes f,L(y\otimes f))z\\ &\quad+\sum_{i=2}^{n-1}p_n(x\otimes f,y\otimes f,\ldots,y\otimes f,\mathop{L(y\otimes f)}_{\substack{\uparrow\\ i\,\text{th}}},y\otimes f,\ldots,y\otimes f)z\\ &=(-1)^{n-1}(y\otimes f)L(x\otimes f)z+p_2(x\otimes f,L(y\otimes f))z+0\\ &=(-1)^{n-1}f(L(x\otimes f)z)y+f(L(y\otimes f)z)x. \end{align*} $$

In an analogous manner, we also have

$$ \begin{align*} L(x\otimes f)z=(-1)^{n-1}f(L(x\otimes f)z)y'+f(L(y'\otimes f)z)x. \end{align*} $$

The linear independence of $y,y',x$ implies that $f(L(x\otimes f)z)=0$ . We therefore have $L(x\otimes f)z\in \mathbb {C} x$ .

If $x\notin \ker f$ , for an arbitrary $z\in \mathcal {X}(\mathcal {A})\cap \ker f$ , it follows that

(3-11) $$ \begin{align} \notag f(x)^{n-1}L(z\otimes f)&=L(p_n(z\otimes f,x\otimes f,\ldots,x\otimes f))\\ &=\sum_{i=1}^{n}p_n(z\otimes f,x\otimes f,\ldots,\mathop{L(x\otimes f)}_{\substack{\uparrow\\ i\,\text{th}}},\ldots,x\otimes f). \end{align} $$

By a direct verification,

$$ \begin{align*} p_n(z\otimes f,x\otimes f,\ldots,&\mathop{L(x\otimes f)}_{\substack{\uparrow\\ i\,\text{th}}},\ldots,x\otimes f)\\ &=f(x)^{i-2}p_{n-(i-2)}(z\otimes f,L(x\otimes f),x\otimes f,\ldots,x\otimes f)\;(i>1) \end{align*} $$

and

$$ \begin{align*}p_n(z\otimes f,x\otimes f,\ldots,x\otimes f,L(x\otimes f))=f(x)^{n-2}p_2(z\otimes f,L(x\otimes f)). \end{align*} $$

For an arbitrary $A_1\in \mathcal {B(X)}$ , one can define $A_i:=p_i(A_1,x\otimes f,\ldots ,x\otimes f)\;(i\in \mathbb {N})$ . By an induction on i, it follows that

(3-12) $$ \begin{align} p_2(A_{k},x\otimes f)x=f(x)^{k}A_1x-\sum_{i=2}^{k}f(x)^{k-i}f(A_ix)x\quad\text{for all } k\in\mathbb{N}. \end{align} $$

Considering the action of (3-11) on x and using (3-12), we obtain

where $\unicode{x3bb} _1,\unicode{x3bb} _2\in \mathbb {C}$ . Simplifying the equality above, we get

This implies that $(L(x\otimes f)-h(x,f)I)z\in \mathbb {C} x$ .

We next show that $h(x,f)$ is bilinear. Consider $f\in \mathcal {X}^\bot _-(\mathcal {A})$ , $x,y\in \mathcal {X}(\mathcal {A})$ and $a,b\in \mathbb {C}$ . If $x,y\in \ker f$ , then this case is trivial. If $x\in \ker f$ while $y\notin \ker f$ , by Condition 4 we can choose $z\in \mathcal {X}(\mathcal {A})\cap \ker f$ such that $x,y,z$ are linearly independent. Considering the actions of $L(x\otimes f)$ , $L(y\otimes f)$ and $L((ax+by)\otimes f)$ on z, the linearity of L entails

$$ \begin{align*}(h(ax+by,f)-bh(y,f))z\in\mathbb{C} x+\mathbb{C} y. \end{align*} $$

Thus $h(ax+by,f)=bh(y,f)$ . If $x,y\notin \ker f$ , and $x,y$ are linearly dependent, then we can find $z\in \mathcal {X}(\mathcal {A})\cap \ker f$ such that $z,x$ are linearly independent. Taking into account the actions of $L(x\otimes f)$ , $L(y\otimes f)$ , $L((ax+by)\otimes f)$ on z, the linearity of L gives

$$ \begin{align*}(h(ax+by,f)-ah(x,f)-bh(y,f))z\in\mathbb{C} x. \end{align*} $$

Hence $h(ax+by,f)=ah(x,f)+bh(y,f)$ . If $x,y\notin \ker f$ are linearly independent, by Condition 4 we can pick $z^\prime \in \mathcal {X}(\mathcal {A})\cap \ker f$ such that the two elements $z^\prime $ and $z:=y-({f(y)}/{f(x)})x\in \mathcal {X}(\mathcal {A})\cap \ker f$ are linearly independent. It is not difficult to see the linear independence of $x,y,z'$ follows from the linear independence of $x,z,z'$ . Using $L(x\otimes f)$ , $L(y\otimes f)$ , and $L((ax+by)\otimes f)$ to act on $z'$ , the linearity of L implies that

$$ \begin{align*} (h(ax+by,f)-ah(x,f)-bh(y,f))z'\in\mathbb{C} x+\mathbb{C} y. \end{align*} $$

Therefore $h(ax+by,f)=ah(x,f)+bh(y,f)$ . In any case, $h(x,f)$ is linear with respect to x.

To see the linearity of $h(x, f)$ with regard to f, let us fix $x \in \mathcal{X}_{-} ^{\bot}(\mathcal{A})$ and take arbitrary $f_1,f_2\in \mathcal {X}^\bot _-(\mathcal {A})$ and $a,b\in \mathbb {C}$ , if $x\in \ker f_1\cap \ker f_2$ , then the case is trivial; if $x\notin \ker f_1\cap \ker f_2$ , by Condition 4 there exists $z\in \mathcal {X}(\mathcal {A})\cap \ker f_1\cap \ker f_2$ such that $z, x$ are linearly independent. Considering the actions of $L(x\otimes f_1)$ , $L(x\otimes f_2)$ , and $L(x\otimes (af_1+bf_2))$ on z, the linearity of L yields

$$ \begin{align*} (h(x,af_1+bf_2)-ah(x,f_1)-bh(x,f_2))z\in\mathbb{C} x. \end{align*} $$

Hence $h(x,af_1+bf_2)=ah(x,f_1)+bh(x,f_2)$ . In any case, $h(x,f)$ is linear with respect to f.

Finally, we prove that if $\mathcal {X}=\mathcal {X}(\mathcal {A})$ and L is continuous, then $h(x,f)$ is continuous for x. Fixing $f\in \mathcal {X}^\bot _-(\mathcal {A})$ and $x_0\in \mathcal {X}(\mathcal {A})\setminus \ker f$ , for an arbitrary sequence $\{x_m\}\subseteq \mathcal {X}$ satisfying $x_m\longrightarrow x_0$ , there exists $M\in \mathbb {N}$ such that for any $m\geqslant M$ , $x_m\notin \ker f$ (by continuity of f). Then for the sequence $\{x_m\}_{m=M}^\infty $ , we have

$$ \begin{align*} h(x_m,f)=\frac{1}{f(x_m)}f(L(x_m\otimes f)x_m)\longrightarrow \frac{1}{f(x_0)}f(L(x_0\otimes f)x_0)=h(x_0,f). \end{align*} $$

Thus $h(x,f)$ is continuous at $x_0$ for a fixed $f\in \mathcal {X}^\bot _-(\mathcal {A})$ . Since $\mathcal {X}(\mathcal {A})=\mathcal {X}$ is a linear space, the linear functional $h_f(x):=h(x,f)$ is continuous. Hence, $h(x,f)$ is continuous in x.

Proof. For any $z\in \mathcal {X}(\mathcal {A})=\mathcal {X}$ , we have $z'=z-f(z)y\in \mathcal {X}(\mathcal {A})\cap \ker f$ . Thus,

$$ \begin{align*} (L(x\otimes f)-h(x,f)I)z&=(L(x\otimes f)-h(x,f)I)(f(z)y+z')\\ &=f(z)\Phi_{f,y}x+\varphi_{x,f,y}(z')x\\ &=f(z)\Phi_{f,y}x+\varphi_{x,f,y}(z)x\\ &=(\Phi_{f,y}x\otimes f+x\otimes \varphi_{x,f,y})z.\\[-35pt] \end{align*} $$

Proof. Let us take $f\in \mathcal {X}^\bot _-(\mathcal {A})\setminus \{0\}$ and $y\in \mathcal {I}(f)$ . For linearly independent $x_1,x_2\in \mathcal {X}\setminus \{0\}$ , we can choose an arbitrary $z\in \ker f$ . By Lemma 3.7, it follows that

$$ \begin{align*} f(z)\Phi_{f,y}(x_1+x_2)+&\varphi_{x_1+x_2,f,y}(z)(x_1+x_2)\\ &=((L(x_1+x_2)\otimes f)-h(x_1+x_2,f)I)z\\ &=(L(x_1\otimes f)-h(x_1,f)I+L(x_2\otimes f)-h(x_2,f)I)z\\ &=f(z)\Phi_{f,y}(x_1)+\varphi_{x_1,f,y}(z)x_1+f(z)\Phi_{f,y}(x_2)+\varphi_{x_2,f,y}(z)x_2, \end{align*} $$

from which we get (note that $\varphi _{x,f,y}$ is an extension of $\varphi _{x,f}$ )

$$ \begin{align*} (\varphi_{x_1+x_2,f}(z)-\varphi_{x_1,f}(z))x_1=(\varphi_{x_2,f}(z)-\varphi_{x_1+x_2,f}(z))x_2. \end{align*} $$

We therefore have $\varphi _{x_1,f}=\varphi _{x_1+x_2,f}=\varphi _{x_2,f}$ . Since $\dim \mathcal {X}^\bot _->1$ , we see that $\dim \mathcal {X}>1$ . For linearly dependent $x_1, x_2\in \mathcal X \setminus \{0\}$ , we can pick $x_3$ such that $x_3$ and $x_1$ are linearly independent. In view of the previous proof, we conclude that $\varphi _{x_1,f}=\varphi _{x_3,f}=\varphi _{x_2,f}$ .

Proof. If $f_1$ and $f_2$ are linearly independent, then by [Reference Kadison and Ringrose14, Proposition 1.1.1], $\ker f_1$ and $\ker f_2$ do not contain each other. Thus there are $x_1,\,x_2\in \mathcal {X}=\mathcal {X}(\mathcal {A})$ such that $f_i(x_j)=\delta _{ij}$ , where $\delta _{ij}$ is the Kronecker delta. For any $x\in \mathcal {X}$ , it follows that

(3-14) $$ \begin{align} &\Phi_{f_1,y_1}x\otimes f_1+x\otimes \varphi_{f_1,y_1}+\Phi_{f_2,y_2}x\otimes f_2+x\otimes\varphi_{f_2,y_2}\nonumber\\ &\quad=L(x\otimes (f_1+f_2))-h(x,f_1+f_2)I\nonumber\\ &\quad=\Phi_{f_1+f_2,y'}x\otimes(f_1+f_2)+x\otimes\varphi_{f_1+f_2,y'}, \end{align} $$

where $y'\in \mathcal {I}(f_1+f_2)$ . Applying (3-14) to $x_1-x_2$ yields

$$ \begin{align*} (\Phi_{f_1,y_1}-\Phi_{f_2,y_2})x=(\varphi_{f_1+f_2,y'}(x_1-x_2)-\varphi_{f_1,y_1}(x_1-x_2)-\varphi_{f_2,y_2}(x_1-x_2))x. \end{align*} $$

If $f_1$ and $f_2$ are linearly dependent, then we can assume $f_1=af_2$ and $a\neq 0$ . It is easy to see that $ay_1\in \mathcal {I}(f_2)$ . We now arrive at

$$ \begin{align*} \Phi_{f_1,y_1}=\Phi_{af_2,y_1}=\Phi_{f_2,ay_1}. \end{align*} $$

For any $y, y'\in \mathcal {I}(f_2)$ , then $y-y'\in \mathcal {X}(\mathcal {A})\cap \ker f_2$ . For any $x\in \mathcal {X}(\mathcal {A})$ , we by Lemma 3.8 get

$$ \begin{align*} (\Phi_{f_2,y}-\Phi_{f_2,y'})x=(L(x\otimes f_2)-h(x,f_2)I)(y-y')=\varphi_{f_2}(y-y') x. \end{align*} $$

Let us put $y=ay_1$ , $y'=y_2$ . Then we obtain $\Phi _{f_1,y_1}-\Phi _{f_2,y_2}\in \mathbb {C}I$ .

Proof. For any $z\in \ker f$ , it follows from (3-13) that

$$ \begin{align*}f(\Phi_{f,y}z)=f(L(z\otimes f)y)\, \, \text{and}\, \, \varphi_{f,y}(z)=\varphi_{y,f,y}(z)=f(L(y\otimes f)z). \end{align*} $$

By invoking Lemma 3.6, we have

$$ \begin{align*} h(y+z,f)&=f(L(y+z)\otimes f)(y+z))\\ &=f(L(y\otimes f)y)+f(\Phi_{f,y}z)+\varphi_{f,y}(z)+0\\ &=h(y,f)+f(\Phi_{f,y}z)+\varphi_{f,y}(z)+0. \end{align*} $$

Thus $f(\Phi _{f,y}z)+\varphi _{f,y}(z)=h(z,f)=0$ . Relations $f(\Phi _{f,y}y)=f(L(y\otimes f)y)-h(y,f)=0$ , $\varphi _{f,y}(y)=0$ and $\ker f+\mathcal {I}(f)= X$ give the proof.

Proof. For any $x\in \mathcal {X}\setminus \{0\}$ , by Lemma 2.1 we know that there exists $f\in \mathcal {X}^\bot _-(\mathcal {A})\setminus \{0\}$ such that $f(x)=0$ . For a fixed $y\in \mathcal {I}(f)$ , let us define a mapping $D(\diamond )=[\,\Phi_{f,y}, \diamond \,]\colon \mathcal {A}\longrightarrow \text { Hom}(\mathcal {X}, \mathcal {X})$ . In view of Lemma 3.9, we can say that D is a linear operator which is unrelated to the choice of $f,y$ (with D bounded and $D\colon \mathcal {A}\longrightarrow \mathcal {B(X)}$ , if L is continuous). Let us first consider the case $n>2$ . Using Lemma 3.7 together with straightforward computation shows that for all $A\in \mathcal {A}$

(3-15) $$ \begin{align} &L(p_n(A,x\otimes f,y\otimes f,\ldots,y\otimes f))\notag\\ \notag&\quad=L(p_{n-1}(Ax\otimes f-x\otimes A^*f,y\otimes f,\ldots,y\otimes f))\\ \notag&\quad=L(Ax\otimes f-f(Ay)x\otimes f-f(Ax)y\otimes f)\\ \notag&\quad=\Phi_{f,y} Ax\otimes f+Ax\otimes\varphi_{f,y}+h(Ax,f)I-f(Ay)(\Phi_{f,y} x\otimes f+x\otimes\varphi_{f,y})\\ &\qquad-f(Ax)(\Phi_{f,y} y\otimes f+y\otimes\varphi_{f,y}+h(y,f)I). \end{align} $$

On the other hand, we have

(3-16) $$ \begin{align} &L(p_n(A,x\otimes f,y\otimes f,\ldots,y\otimes f))\nonumber\\ &\quad=p_n(L(A),x\otimes f,y\otimes f,\ldots,y\otimes f)+p_n(A,L(x\otimes f),y\otimes f,\ldots,y\otimes f)\notag\\ &\qquad+\sum_{j=3}^np_n(A,x\otimes f,\ldots,\mathop{L(y\otimes f)}_{\substack{\uparrow\\ j\,\text{th}}},\ldots,y\otimes f). \end{align} $$

We calculate the terms of (3-16) in turn with the help of Lemma 3.10. Let us see the first term in (3-16):

$$ \begin{align*} p_n(L(A),x\otimes f,y\otimes f,\ldots, y\otimes f)&=p_{n-1}(L(A)x\otimes f-x\otimes L(A)^*f,y\otimes f,\ldots,y\otimes f)\\ &=L(A)x\otimes f-f(L(A)y)x\otimes f-f(L(A)x)y\otimes f. \end{align*} $$

The second term in (3-16) is

$$ \begin{align*} &p_n(A,L(x\otimes f),y\otimes f,\ldots,y\otimes f)\\ &\quad =p_n(A,\Phi_{f,y} x\otimes f,y\otimes f,\ldots,y\otimes f)+p_n(A,x\otimes\varphi_{f,y},y\otimes f,\ldots,y\otimes f)\\ &\quad =\left\{\begin{array}{@{}ll} A\Phi_{f,y} x\otimes f-f(Ay)\Phi_{f,y} x\otimes f-f(A\Phi_{f,y} x)y\otimes f&\\ +f(\Phi_{f,y}x)y\otimes A^*f-\varphi_{f,y}(Ay)x\otimes f-f(Ax)y\otimes\varphi_{f,y},&\;(n \text{ is odd})\\ & \\ A\Phi_{f,y} x\otimes f-f(Ay)\Phi_{f,y} x\otimes f-f(A\Phi_{f,y} x)y\otimes f&\\ +2f(Ay)f(\Phi_{f,y}x)y\otimes f-f(\Phi_{f,y}x)y\otimes A^*f&\\ -\varphi_{f,y}(Ay)x\otimes f+f(Ax)y\otimes\varphi_{f,y},&\;(n \text{ is even}). \end{array}\right. \end{align*} $$

The remaining terms in $\sum $ (for the case of $3\leq j\leq n$ ) of (3-16) are

$$ \begin{align*} &p_n(A,x\otimes f,\ldots,\mathop{L(y\otimes f)}_{\substack{\uparrow\\ j\,\text{th}}},\ldots,y\otimes f)\\ &\quad=p_n(A,x\otimes f,\ldots,\mathop{\Phi_{f,y} y\otimes f} _{\substack{\uparrow\\ j\,\text{th}}},\ldots,y\otimes f)+p_n(A,x\otimes f,\ldots,\mathop{y\otimes\varphi_{f,y}} _{\substack{\uparrow\\ j\,\text{th}}},\ldots,y\otimes f)\\ &\quad=\left\{\begin{array}{@{}ll} -f(A\Phi_{f,y} y)x\otimes f-f(Ax)\Phi_{f,y} y\otimes f+Ax\otimes\varphi_{f,y}\\ -f(Ay)x\otimes\varphi_{f,y}-\varphi_{f,y}(Ax)y\otimes f+\varphi_{f,y}(x)y\otimes A^*f, &n=3,\\ & \\ \left\{\begin{array}{@{}ll} j=3,\;&-f(A\Phi_{f,y} y)x\otimes f-f(Ax)\Phi_{f,y} y\otimes f\\ &+\varphi_{f,y}(x)f(Ay)y\otimes f-f(Ax)y\otimes \varphi_{f,y}\\ &-\varphi_{f,y}y\otimes A^*f,\\ 3<j<n,\;& 0, \end{array}\right. &n>3,\;n \text{ is odd},\\ & \\ \left\{\begin{array}{@{}ll} j=3,\;&-f(A\Phi_{f,y} y)x\otimes f-f(Ax)\Phi_{f,y} y\otimes f\\ &-\varphi_{f,y}(x)f(Ay)y\otimes f+f(Ax)y\otimes \varphi_{f,y}\\ &+\varphi_{f,y}(x)y\otimes A^*f,\\ 3<j<n,\;& 0, \end{array}\right. &n>3,\;n \text{ is even},\\ & \\ Ax\otimes\varphi_{f,y}-f(Ay)x\otimes \varphi_{f,y}-f(Ax)y\otimes \varphi_{f,y}\\ -\varphi_{f,y}(Ax)y\otimes f+f(Ay)\varphi_{f,y}(x)y\otimes f. &n>3,\;j=n, \end{array}\right. \end{align*} $$

Combining (3-15) with (3-16) and considering their actions on x gives

$$ \begin{align*} h(Ax,f)x-f(Ax)h(y,f)x\in\mathbb{C} y. \end{align*} $$

Thus $h(Ax,f)-f(Ax)h(y,f)=0$ .

For n odd with $n>3$ , using the relations (3-15) and (3-16) to act on y, we arrive at

$$ \begin{align*} (L(A)&+A\Phi_{f,y}-\Phi_{f,y}A)x-f(L(A)y)x-f(L(A)x)y-f(A\Phi_{f,y}x)y \\[-2pt] &+f(\Phi_{f,y}x)f(Ay)y-\varphi_{f,y}(Ay)x+\varphi_{f,y}(x)f(Ay)y-\varphi_{f,y}(x)f(Ay)y\\[-2pt] &-\varphi_{f,y}(Ax)y+f(Ay)\varphi_{f,y}(x)y=0. \end{align*} $$

For n even with $n>3$ , from the actions of (3-15) and (3-16) on y,

$$ \begin{align*} (L(A)&+A\Phi_{f,y}-\Phi_{f,y}A)x-f(L(A)y)x-f(L(A)x)y-f(A\Phi_{f,y}x)y\\[-2pt] &+2f(Ay)f(\Phi_{f,y}x)y-f(\Phi_{f,y}x)f(Ay)y-\varphi_{f,y}(Ay)x-\varphi_{f,y}(x)f(Ay)y\\[-2pt] &+\varphi_{f,y}(x)f(Ay)y-\varphi_{f,y}(Ax)y+f(Ay)\varphi_{f,y}(x)y=0. \end{align*} $$

For the case of $n=3$ , adopting an analogous calculation we see that

$$ \begin{align*} (L(A)&+A\Phi_{f,y}-\Phi_{f,y}A)x-f(L(A)y)x-f(L(A)x)y-f(A\Phi_{f,y}x)y\\[-2pt] &+f(\Phi_{f,y}x)f(Ay)y-\varphi_{f,y}(Ay)x-f(A\Phi_{f,y}y)x-\varphi_{f,y}(Ax)y\\[-2pt] &+\varphi_{f,y}(x)f(Ay)y=0. \end{align*} $$

Therefore, for $n\geqslant 3$ , there exist $a_{A,f,y},\;b_{A,f,y}\in \mathbb {C}$ such that

(3-17) $$ \begin{align} (L(A)+A\Phi_{f,y}-\Phi_{f,y}A)x+a_{A,f,y}x+b_{A,f,y}y=0. \end{align} $$

Since $\dim \mathcal {X}>2$ , we can say that $\dim \ker f>1$ . Thus there exists $z\in \ker f$ such that $z,x$ are linearly independent. Then $x,\;y,\;y+z$ are linearly independent. Indeed, if $ax+by+c(y+z)=0$ ( $a,b,c\in \mathbb {C}$ ), then $ax+(b+c)y+cz=0$ . Now $y\notin \ker f$ , and hence $b+c=0$ . Since $x, z$ are linearly independent, it follows that $a=b=c=0$ . Note that $y+z\in \mathcal {I}(f)$ . Repeating the above argument gives

(3-18) $$ \begin{align} (L(A)+A\Phi_{f,y+z}-\Phi_{f,y+z}A)x+a_{A,f,y+z}x+b_{A,f,y+z}(y+z)=0. \end{align} $$

Note that $D(A)=[\Phi _{f,y},A]=[\Phi _{f,y+z},A]$ . By (3-17) and (3-18), we conclude that

$$ \begin{align*}(a_{A,f,y}-a_{A,f,y+z})x+b_{A,f,y}y-b_{A,f,y+z}(y+z)=0, \end{align*} $$

which entails $b_{A,f,y}=0$ . It follows from (3-17) that $(L(A)-D(A))x\in \ker f$ . For an arbitrary $y'\in \mathcal {I}(f)$ , applying f to

$$ \begin{align*} (L(A)+A\Phi_{f,y'}-\Phi_{f,y'}A)x+a_{A,f,y'}x+b_{A,f,y'}y'=0 \end{align*} $$

yields $b_{A,f,y'}=0$ . So $(L(A)-D(A))x=-a_{A,f,y'}x$ . The arbitrariness of $y'$ implies that $a_{A,f,y}$ is unrelated to the choice of $y\in \mathcal {I}(f)$ . Let us denote it by $a_{A,f}$ . Similarly, changing $f\in \mathcal {X}^\bot _-(\mathcal {A})\setminus \{0\}$ satisfying the condition $f(x)=0$ , we see that $a_{A,f}$ is merely related to the choice of x. Thus we may write $H_x(A):=-a_{A,f}$ . Therefore, $\text{for all } x\in \mathcal{X}$ ,

(3-19) $$ \begin{align} L(A)x=D(A)x+H_x(A)x. \end{align} $$

For the case of $n=2$ , if $A^*f\neq 0$ , fixing $y'\in \mathcal {I}(A^*f)$ , by Lemma 3.9 we see that $\Phi _{A^*f,y'}-\Phi _{f,y}=cI$ . Let us put $\psi _{A^*f}=\varphi _{f,y}+cf$ . We calculate

$$ \begin{align*} L(p_2(A,x\otimes f))&=L(Ax\otimes f-x\otimes A^*f)\nonumber\\ &=\Phi_{f,y} Ax\otimes f+Ax\otimes\varphi_{f,y}+h(Ax,f)I\nonumber\\ &\quad-\Phi_{f,y} x\otimes A^*f-x\otimes\psi_{A^*f}-h(x,A^*f)I \end{align*} $$

and

$$ \begin{align*} L(p_2(A,x\otimes f))&=p_2(L(A),x\otimes f)+p_2(A,L(x\otimes f))\\ &=L(A)x\otimes f-x\otimes L(A)^*f+A\Phi_{f,y} x\otimes f\\ &\quad+Ax\otimes\varphi_{f,y}-\Phi_{f,y} x\otimes A^*f-x\otimes A^*\varphi_{f,y}. \end{align*} $$

Thus we arrive at

(3-20) $$ \begin{align} (L(A)+A\Phi_{f,y}-\Phi_{f,y} A)x\otimes f&=x\otimes L(A)^*f+x\otimes A^*\varphi_{f,y}+h(Ax,f)I\nonumber\\ &\quad-x\otimes \psi_{A^*f}-h(x,A^*f)I. \end{align} $$

In an analogous manner, we also have (writing $y_1=y$ , $y_2=y+x$ )

(3-21) $$ \begin{align} (L(A)+A\Phi_{f,y}-\Phi_{f,y} A)y_i\otimes f&=y_i\otimes L(A)^*f+y_i\otimes A^*\varphi_{f,y}+h(Ay_i,f)I\nonumber\\ &\quad-y_i\otimes \psi_{A^*f}-h(y_i,A^*f)I. \end{align} $$

If $A^*f=0$ , by a similar argument, we obtain (3-20) and (3-21) if we substitute $0$ for $\psi _{A^*f}$ and $h(\diamond ,A^*f)$ . Applying (3-20) to y yields

(3-22) $$ \begin{align} (L(A)+A\Phi_{f,y}-\Phi_{f,y} A)x&=(f(L(A)y)+\varphi_{f,y}(Ay)-\psi_{A^*f}(y)) x\nonumber\\ &\quad+(h(Ax,f)-h(x,A^*f))y. \end{align} $$

Applying (3-21) to x gives

$$ \begin{align*} 0=(f(L(A)x)+\varphi_{f,y}(Ax)-\psi_{A^*f}(x))y_i+(h(Ay_i,f)-h(y_i,A^*f))x. \end{align*} $$

It is not difficult to see that x and $y_i\ (i=1,2)$ are linearly independent. So $h(Ay_1,f)-h(y_1,A^*f)=0$ and $h(Ay_2,f)-h(y_2,A^*f)=0$ . Their subtraction gives $h(Ax,f)-h(x,A^*f)=0$ . By (3-22), $f(L(A)y)+\varphi _{f,y}(Ay)-\psi _{A^*f}(y)=-a_{A,f}$ depends only on the choice of x and A; let us denote it by $H_x(A)$ . Therefore, for the case of $n=2$ , the relation (3-19) still holds true for any $x\in \mathcal {X}$ . That is,

$$ \begin{align*}\begin{aligned} L(A)x=[\Phi_{f, y}, A]x+H_x(A)x \quad\text{for all } x\in \mathcal{X}. \end{aligned} \end{align*} $$

Finally, for $x_j\in \mathcal {X}\setminus \{0\} \ (\kern1.5pt j=1,2)$ and $f_j\in \mathcal {X}^\bot _-(\mathcal {A})\setminus \{0\}$ satisfying the condition $f_j(x_j)=0$ (by Lemma 2.1 there are such $f_j$ ), since the codimension of $\ker f_1\cap \ker f_2$ is no more than $2$ , there is $z\in (\ker f_1\cap \ker f_2)\setminus \{0\}$ . According to the preceding discussion, we obtain that $H_{x_1}(A)=-a_{A,f_1}=H_z(A)=-a_{A,f_2}=H_{x_2}(A)$ . This shows that $H_x(A)$ is unrelated to the choice of x. One can establish a mapping $H\colon \mathcal {A}\longrightarrow \mathbb {C}I$ by $H(A)x:=H_x(A)x\;(\text {for all } x\in \mathcal {X})$ . In light of (3-19), we know that H is linear (and continuous if L is continuous). Then by (3-19) again, one can see $D=L-H$ , which implies $D\colon \mathcal {A}\longrightarrow \mathcal {B(X)}$ . It follows from Proposition 2.5 that H vanishes on all $(n-1)$ commutators on $\mathcal {A}$ . Thus $L=D+H$ is the desired standard form.

Proof. Note that if $\mathcal {A}=\text {Alg}\mathcal {N}$ , then $\mathcal {X}^\bot _-(\mathcal {A})=\mathcal {X}^\bot _-$ . If $\text {codim}\, \mathcal {X}_->1$ , then by the Hahn–Banach theorem one can construct $x_1,x_2\in \mathcal X\setminus \mathcal X_-$ and $f_1,f_2\in \mathcal {X}^\bot _-$ such that $f_i(x_j)=\delta _{ij}$ , where $\delta _{ij}$ is the Kronecker delta. Thus $f_1$ and $f_2$ are linearly independent, which entails $\dim \mathcal {X}^\bot _->1$ . Then by Theorem 3.11 and Lemma 2.8, we immediately get the conclusion.

If $\text {codim}\,\mathcal {X}_-=1$ , we see that $\text {codim}\,\ker f=1$ and $\mathcal X_-\subseteq \ker f$ for any $f\in \mathcal {X}^\bot _-\setminus \{0\}$ . By Lemma 2.2, it follows that $\mathcal {X}_-=\ker f$ . Taking into account [Reference Kadison and Ringrose14, Proposition 1.1.1], we get $\dim \mathcal {X}^\bot _-=1$ . In view of Theorem 3.3, we complete the proof.

Proof. It follows from the assumption $\mathcal {X}_-\subsetneqq \mathcal {X}$ that $\{\mathbf 0\}=\mathcal {X}^\bot \subsetneqq \mathcal {X}^\bot _-$ , in which the strict inclusion relation is given by the Hahn–Banach theorem. For any $f\in \mathcal {X}^*$ , we have

$$ \begin{align*}\begin{aligned} f\in\mathcal{X}^\bot_- & \Leftrightarrow\text{ for all } x\in N\subsetneqq \mathcal{X},\,f(x)=0\\ & \Leftrightarrow\text{ for all } N\subsetneqq \mathcal{X},\, N\subseteq\ker f\\ &\Leftrightarrow\text{ for all } N\subsetneqq \mathcal{X},\,f\in N^\bot\\ &\Leftrightarrow f\in\{\mathbf 0\}_+. \end{aligned} \end{align*} $$

Thus $\mathcal {X}^\bot _-=\{\mathbf 0\}_+$ , and so $\{\mathbf 0\}\subsetneqq \{\mathbf 0\}_+$ .

Conversely, $f\in \{\mathbf 0\}_+\setminus \{\mathbf 0\}$ ; then for any $x\in N\subsetneqq \mathcal {X}$ , we have $f(x)=0$ . The continuity of f implies that $f(\bigvee _{N\subsetneqq \mathcal {X}}N)=\{0\}$ . Since $f\neq 0$ , we get $f(\mathcal {X})\neq \{0\}$ . Therefore $\mathcal {X}_-\subsetneqq \mathcal {X}$ .

Proof. The linearity of $L^*$ is straightforward. If L is continuous, since the Banach adjoint $*$ is continuous, the continuity of $L^*$ follows. For arbitrary elements $A_1,\ldots ,A_n\in \text {Alg}\mathcal {N}$ , we have

$$ \begin{align*} L^*(p_n(A_1^*,\ldots,A_n^*))&=(-1)^{n-1}L^*(p_n(A_1,\ldots,A_n)^*)\\ &=(-1)^{n-1}L(p_n(A_1,\ldots,A_n))^*\\ &=(-1)^{n-1}\bigg(\sum_{i=1}^np_n(A_1,\ldots,L(A_i),\ldots,A_n)\bigg)^*\\ &=(-1)^{n-1}\sum_{i=1}^np_n(A_1,\ldots,L(A_i),\ldots,A_n)^*\\ &=\sum_{i=1}^np_n(A_1^*,\ldots,L^*(A_i^*),\ldots,A_n^*). \end{align*} $$

This shows that $L^*$ is a Lie n-derivation.

Proof. Let us write $\mathcal {A}=(\text {Alg}\mathcal {N})^*$ . As is demonstrated above, we know that $\dim ((\mathcal {X}^*)^\bot _-(\mathcal {A}))\geqslant 1$ . It follows from Theorem 3.3 and Theorem 3.11 that the Lie n-derivation $L^*$ has the standard form, that is, there is a derivation $D'\colon \mathcal {A}\longrightarrow \mathcal {B}(\mathcal {X}^*)$ and linear mapping $H'\colon \mathcal {A}\longrightarrow \mathbb {C}I^*$ vanishing on all $(n-1)\,$ th commutators on $(\text {Alg}\mathcal {N})^*$ such that $L^*=D'+H^\prime $ (by Lemma 3.15, $D^\prime $ and $H^\prime $ are continuous if L is continuous). Let us define $H(A)^*:=H'(A^*)$ . It is easy to check that $H\colon \text {Alg}\mathcal {N}\longrightarrow \mathbb {C}I$ is a linear mapping (continuous whenever L is continuous). For any $A,B\in \text {Alg}\mathcal {N}$ , we get

$$ \begin{align*} ((L-H)(AB))^*&=(L^*-H')(B^*A^*)\\ &=D'(B^*A^*)=D'(B^*)A^*+B^*D'(A^*)\\ &=((L-H)(B))^*A^*+B^*((L-H)(A))^*\\ &=((L-H)(A)B+A(L-H)(B))^*. \end{align*} $$

Since $*$ is isometric, we know that

$$ \begin{align*}(L-H)(AB)=(L-H)(A)B+A(L-H)(B). \end{align*} $$

Thus $D=L-H$ is a derivation (and is continuous if L is continuous).

Proof. (i) Suppose that $N_0\in \mathcal {N}\setminus \{\{0\}, \mathcal {X}\}$ is of finite dimension. By the fact $\{0\}=\{0\}_+=\bigwedge \{N\in \mathcal {N}\mid \{0\}\subsetneqq N\}$ and the total-ordering of $\mathcal {N}$ , it follows that there exists $m\in \mathbb {N}$ and $N_1,\ldots ,N_m\in \mathcal {N}\setminus \{\{0\}\}$ with $N_j\subseteq N_0$ , such that $\{0\}=\bigwedge _{j=0}^n N_j$ , while $\dim \bigwedge _{j=0}^n N_j\geqslant 1$ , which is a contradiction.

Suppose that $N_0\in \mathcal {N}{\setminus} \{\{0\}, \mathcal {X}\}$ has finite codimension. Then $\{N\in \mathcal {N}\mid N_0\subsetneqq N\subsetneqq \mathcal {X}\}$ is a finite set. Indeed, for $N_0\subsetneqq N$ , since $\mathcal {X}/N\cong (\mathcal {X}/N_0)/(N/N_0)$ , there are at most $\dim \mathcal {X}/N_0=\text {codim}\, N_0<\infty $ possible choices for $\dim \mathcal {X}/N$ . Moreover, since $\mathcal {N}$ is totally ordered, we see that there are at most codim $N_0$ choices for $\mathcal {X}/N$ . Let us now write $\{N\in \mathcal {N}\mid N_0\subsetneqq N\subsetneqq \mathcal {X}\}=\{N_j\}_{j=0}^m$ , which is an increasing sequence under the inclusion relation. Then

$$ \begin{align*}\mathcal{X}_-=\overline{\text{span}\,\bigcup\{N\in\mathcal{N}\mid N\subsetneqq \mathcal{X}\}}=\overline{\text{span}\,\bigcup_{j=0}^mN_j}=N_m\subsetneqq \mathcal{X}, \end{align*} $$

which contradicts that $\mathcal {X}_-=\mathcal {X}$ .

(ii) If $N\subseteq \ker f$ , then by (1) we have $\dim N\cap \ker f=\dim N=\infty $ . If $N\not \subseteq \ker f$ , then there exists $y\in N$ such that $f(y)=1$ . Thus for any $x\in N$ , $x-f(x)y\in \ker f$ ; that is, $(I-y\otimes f)\mathcal {X}\subseteq \ker f$ . Therefore $(I-y\otimes f)N\subseteq N\cap \ker f$ , where $(I-y\otimes f)N$ is an infinite-dimensional subspace. Henceforth we get $\dim N\cap \ker f=\infty $ .

(iii) If $N_{\unicode{x3bb} }\subseteq \ker f$ holds for every $\unicode{x3bb} \in \mathit {\Lambda }$ , then $\bigvee _{\unicode{x3bb} \in \mathit {\Lambda }}N_{\unicode{x3bb} }=\mathcal {X}$ . This implies that $\ker f=\mathcal {X}$ , which is a contradiction.

Proof. For any $x\in N_{\unicode{x3bb} }$ and $f\in (N_{\unicode{x3bb} })_-^\bot $ , by Lemma 3.17 we know that $\dim (N_{\unicode{x3bb} })_-^\bot =\infty $ . Then there exists $f_1\in (N_{\unicode{x3bb} })_-^\bot $ such that $f_1, f$ are linearly independent. By [Reference Kadison and Ringrose14, Proposition 1.1.1], there exists $y_1\in \mathcal {X}$ such that $f(y_1)=0$ and $f_1(y_1)=1$ . For any $A\in \text {Alg}\mathcal {N}$ and $z\in N_{\unicode{x3bb} }\cap \ker f$ , if $n>2$ , it follows that

$$ \begin{align*} 0&=L(p_n(x\otimes f,z\otimes f_1,A,\ldots,A))\\ &=p_n(L(x\otimes f),z\otimes f_1,A,\ldots,A)+p_n(x\otimes f,L(z\otimes f_1),A,\ldots,A)\\ &=p_{n-1}([L(x\otimes f),z\otimes f_1]+[x\otimes f,L(z\otimes f_1)],A\,\ldots,A). \end{align*} $$

By invoking Lemma 2.6 and Lemma 2.8 we deduce that

$$ \begin{align*}[L(x\otimes f),z\otimes f_1]+[x\otimes f,L(z\otimes f_1)]=cI, \end{align*} $$

where $c\in \mathbb {C}$ . Since I is of infinite rank, we have $c=0$ . Thus

(3-23) $$ \begin{align} [L(x\otimes f),z\otimes f_1]+[x\otimes f,L(z\otimes f_1)]=0. \end{align} $$

If $n=2$ , (3-23) can be obtained by expanding $0=L(p_2(x\otimes f,z\otimes f_1))$ . Considering the action of (3-23) on $y_1$ ,

$$ \begin{align*} L(x\otimes f)z=f_1(L(x\otimes f)y_1)z-f(L(z\otimes f_1)y_1)x. \end{align*} $$

We next demonstrate that the equality $f_1(L(x\otimes f)y_1)=f_2(L(x\otimes f)y_2)$ holds true, where $f_j\in (N_{\unicode{x3bb} })_-^\bot $ is linearly independent of f and $y_j\in \mathcal {I}(f)\cap \ker f_j \ (\kern1.5pt j=1,2)$ . By the preceding discussion,

(3-24) $$ \begin{align} L(x\otimes f)z=f_j(L(x\otimes f)y_j)z-f(L(z\otimes f_j)y_j)x\quad(\kern1.5pt j=1,2). \end{align} $$

Since $N_{\unicode{x3bb} }\cap \ker f$ is of infinite dimension, we can take $z'\in N_{\unicode{x3bb} }\cap \ker f$ such that $z', x$ are linearly independent. Similarly,

(3-25) $$ \begin{align} L(x\otimes f)z'=f_j(L(x\otimes f)y_j)z'-f(L(z'\otimes f_j)y_j)x\quad(\kern1.5pt j=1,2). \end{align} $$

Then by linear independence, the two equalities (3-24) and (3-25) imply $f_1(L(x\otimes f)y_1)= f_2(L(x\otimes f)y_2)$ . Let us define $h_{\unicode{x3bb} }(x,f)=f_1(L(x\otimes f)y_1)$ , which is well defined by the above deduction. It is clear that $h_{\unicode{x3bb} }(x,f)$ is continuous for $x\in N_{\unicode{x3bb} }$ if L is continuous.

Let us show the bilinearity of $h_{\unicode{x3bb} }$ . In view of the above construction, linearity of $h_{\unicode{x3bb} }(x,f)$ with respect to x is straightforward to check. For any $f_1,f_2\in (N_{\unicode{x3bb} })_-^\bot $ , if $f_1$ and $f_2$ are linearly dependent, the result is trivial; if $f_1$ and $f_2$ are linearly independent, it follows from the fact $\dim (N_{\unicode{x3bb} })_-^\bot =\infty $ that there exists $f_{12}\in (N_{\unicode{x3bb} })_-^\bot $ such that $f_{12},f_1,f_2$ are linearly independent. By [Reference Kadison and Ringrose14, Proposition 1.1.1], $\ker f_1\cap \ker f_2\not \subseteq \ker f_{12}$ . Then there exists $y_{12}\in \mathcal {I}(f_{12})\cap \ker f_1\cap \ker f_2$ , and so

$$ \begin{align*} h_{\unicode{x3bb}}(x,f_1+f_2)&=f_{12}(L(x\otimes(f_1+f_2))y_{12})\\ &=f_{12}(L(x\otimes f_1)y_{12})+f_{12}(L(x\otimes f)y_{12})\\ &=h_{\unicode{x3bb}}(x,f_1)+h_{\unicode{x3bb}}(x,f_2). \end{align*} $$

That is, $h_{\unicode{x3bb} }(x,f)$ is linear with regard to f as well.

Proof. The case of $n=3$ is given in [Reference Sun and Ma32]. It is sufficient for us to handle the cases of $n>3$ and $n=2$ . Let us first deal with the case of $n>3$ . For an arbitrary $f\in N_{\unicode{x3bb} }^\bot \setminus \{0\}$ , it follows from Lemma 3.17 that there exists $\alpha \in \mathit {\Lambda }$ such that $N_\alpha \not \subseteq \ker f$ . Hence, there exists $y\in \mathcal {I}(f)\cap N_\alpha $ . The fact that $\mathcal {N}$ is totally ordered implies that $N_{\unicode{x3bb} }\subseteq N_\alpha $ . Then, for an arbitrary $g\in N_\alpha ^\bot \setminus \{0\}$ , f and g are linearly independent. By [Reference Kadison and Ringrose14, Proposition 1.1.1], we know that there exists $z\in \mathcal {I}(g)\cap \ker f$ .

Given $f'\in N_{\unicode{x3bb} }^\bot \setminus \{0\}$ and $y'\in \mathcal {I}(f)$ , by Lemma 3.19, $\Phi _{\unicode{x3bb} ,f,y}-\Phi _{\unicode{x3bb} ,f',y'}=c_{\unicode{x3bb} ,f',y'}I|_{N_{\unicode{x3bb} }}$ , and we define $\psi _{\unicode{x3bb} ,f'}:=\varphi _{\unicode{x3bb} ,f,y}+c_{\unicode{x3bb} ,f',y'}I$ , where $c_{\unicode{x3bb} ,f',y'}=0$ if $f'=f$ , $y'=y$ . Then by Lemma 3.19, we get

(3-26) $$ \begin{align} L(x\otimes f')=\Phi_{\unicode{x3bb},f,y} x\otimes f'+x\otimes \psi_{\unicode{x3bb},f'}+h_{\unicode{x3bb}}(x,f')I. \end{align} $$

With the above notation, for arbitrary $x\in N_{\unicode{x3bb} }$ , direct computation shows that

(3-27) $$ \begin{align} &L(p_n(A,x\otimes f,y\otimes f,\ldots,y\otimes f,y\otimes g))\nonumber\\ \notag&\quad=L(p_{n-1}(Ax\otimes f-x\otimes A^*f,y\otimes f,\ldots,y\otimes f,y\otimes g))\\ \notag&\quad=L(Ax\otimes g-f(Ay)x\otimes g)\\ \notag&\quad=\Phi_{\unicode{x3bb},f,y} Ax\otimes g+Ax\otimes\psi_{\unicode{x3bb},g}+h_{\unicode{x3bb}}(Ax,g)I-f(Ay)\Phi_{\unicode{x3bb},f,y} x\otimes g\\ &\qquad-f(Ay)x\otimes \psi_{\unicode{x3bb},g}-f(Ay)h_{\unicode{x3bb}}(x,g)I. \end{align} $$

On the other hand, we have

(3-28) $$ \begin{align} &L(p_n(A,x\otimes f,y\otimes f,\ldots,y\otimes f,y\otimes g))\nonumber\\ \notag&\quad=p_n(L(A),x\otimes f,y\otimes f,\ldots,y\otimes f,y\otimes g)\\ \notag&\qquad+p_n(A,L(x\otimes f),y\otimes f,\ldots,y\otimes f,y\otimes g)\\ \notag&\qquad+\sum_{j=3}^{n-1}p_n(A, x\otimes f,\ldots,\mathop{L(y_j\otimes f)}_{\substack{\uparrow\\ j\,\text{th}}},\ldots,y\otimes g)\\ &\qquad+p_n(A,x\otimes f,y\otimes f,\ldots,y\otimes f,L(y\otimes g)). \end{align} $$

Let us calculate each term on the right side of (3-28). The first and second terms on the right side of (3-28) are

$$ \begin{align*} &p_n(L(A),x\otimes f,y\otimes f,\ldots,y\otimes f,y\otimes g)\\ &\quad=L(A)x\otimes g-f(L(A)y)x\otimes g-f(L(A)x)y\otimes g-g(L(A)x)y\otimes f \end{align*} $$

and

$$ \begin{align*} &p_n(A,L(x\otimes f),y\otimes f,\ldots,y\otimes f,y\otimes g)\\ &\quad=A\Phi_{\unicode{x3bb},f,y} x\otimes g-f(Ay)\Phi_{\unicode{x3bb},f,y} x\otimes g-f(A\Phi_{\unicode{x3bb},f,y} x)y\otimes g+f(\Phi_{\unicode{x3bb},f,y} x)f(Ay)y\otimes g\\ &\qquad-g(A\Phi_{\unicode{x3bb},f,y} x)y\otimes f+f(Ay)g(\Phi_{\unicode{x3bb},f,y} x)y\otimes f-\varphi_{\unicode{x3bb},f,y}(Ay)x\otimes g. \end{align*} $$

The terms in $\sum $ for $3\leqslant j\leqslant n-1$ are

$$ \begin{align*} &p_n(A,x\otimes f,\ldots,\mathop{L(y\otimes f)}_{\substack{\uparrow\\ j\,\text{th}}},\ldots,y\otimes g)\\ &\quad=\left\{ \begin{array}{@{}ll} f(\Phi_{\unicode{x3bb},f,y} y)Ax\otimes g-f(\Phi_{\unicode{x3bb},f,y} y)f(Ay)x\otimes g-\varphi_{\unicode{x3bb},f,y}(Ax)y\otimes g\\ +\varphi_{\unicode{x3bb},f,y}(x)f(Ay)y\otimes g & j=n-1, \\ & \\ f(\Phi_{\unicode{x3bb},f,y} y)Ax\otimes g-f(\Phi_{\unicode{x3bb},f,y} y)f(Ay)x\otimes g & 3<j<n-1, \\ & \\ \left\{ \begin{array}{@{}ll} n=4,&f(\Phi_{\unicode{x3bb},f,y} y)Ax\otimes g-f(A\Phi_{\unicode{x3bb},f,y} y)x\otimes g\\ &-\varphi_{\unicode{x3bb},f,y}(Ax)y\otimes g+\varphi_{\unicode{x3bb},f,y}(x)f(Ay)y\otimes g\\ & \\ n>4,&\left\{ \begin{array}{@{}ll} n\text{ is odd},&f(\Phi_{\unicode{x3bb},f,y} y)Ax\otimes g\\ &-f(A\Phi_{\unicode{x3bb},f,y} y)x\otimes g\\ & \\ n\text{ is even},&2\varphi_{\unicode{x3bb},f,y}(x)f(Ay)y\otimes g\\ &+f(\Phi_{\unicode{x3bb},f,y} y)Ax\otimes g\\ &-f(A\Phi_{\unicode{x3bb},f,y} y)x\otimes g \end{array}\right. \end{array}\right. & j=3, \end{array}\right. \end{align*} $$

The last term is

$$ \begin{align*} &p_n(A,x\otimes f,y\otimes f,\ldots,y\otimes f,L(y\otimes g))\\ &\quad=p_n(A,x\otimes f,y\otimes f,\ldots,y\otimes f,\Phi_{\unicode{x3bb},f,y} y\otimes g+y\otimes \psi_{\lg})\\ &\quad=p_2(Ax\otimes f-f(Ay)x\otimes f-f(Ax)y\otimes f,\Phi_{\unicode{x3bb},f,y} y\otimes g+y\otimes \psi_{\unicode{x3bb},g})\\ &\quad=f(\Phi_{\unicode{x3bb},f,y} y)Ax\otimes g-f(\Phi_{\unicode{x3bb},f,y} y)f(Ay)x\otimes g-f(Ax)f(\Phi_{\unicode{x3bb},f,y} y)y\otimes g+Ax\otimes \psi_{\unicode{x3bb},g}\\ &\qquad-f(Ay)x\otimes \psi_{\unicode{x3bb},g}-\psi_{\unicode{x3bb},g}(Ax)y\otimes f+f(Ay)\psi_{\unicode{x3bb},g}(x)y\otimes f. \end{align*} $$

Comparing (3-27) with (3-28), noticing that I is of infinite rank, we deduce that $h_{\unicode{x3bb} }(Ax,g)-f(Ay)h_{\unicode{x3bb} }(x,g)=0$ . Combining (3-27) with (3-28) and considering their actions to z yields

(3-29) $$ \begin{align} 0&=L(A)x+A\Phi_{\unicode{x3bb},f,y} x-\Phi_{\unicode{x3bb},f,y} Ax-f(L(A)y)x-f(L(A)x)y\notag\\ \notag&\quad-f(A\Phi_{\unicode{x3bb},f,y} x)y+f(\Phi_{\unicode{x3bb},f,y} x)f(Ay)y+f(\Phi_{\unicode{x3bb},f,y} y)Ax-f(\Phi_{\unicode{x3bb},f,y} y)f(Ay)x\\ \notag&\quad-f(Ax)f(\Phi_{\unicode{x3bb},f,y} y)y+(n-4)(f(\Phi_{\unicode{x3bb},f,y} y)Ax-f(\Phi_{\unicode{x3bb},f,y} y)f(Ay)x)\\ \notag&\quad+f(\Phi_{\unicode{x3bb},f,y} y)Ax-f(\Phi_{\unicode{x3bb},f,y} y)f(Ay)x+\varphi_{\unicode{x3bb},f,y}(x)f(Ay)y-\varphi_{\unicode{x3bb},f,y}(Ax)y\\ &\quad+\chi_{(4,\infty)}(n)\cdot\left\{ \begin{array}{@{}ll} f(\Phi_{\unicode{x3bb},f,y} y)Ax-f(A\Phi_{\unicode{x3bb},f,y} y)x & n\text{ is odd}, \\ 2\varphi_{\unicode{x3bb},f,y}(x)f(Ay)y+f(\Phi_{\unicode{x3bb},f,y} y)Ax-f(A\Phi_{\unicode{x3bb},f,y} y)x & n\text{ is even}, \end{array},\right. \end{align} $$

where $\chi _{(4,\infty )}(n)= 1\ \text{or}\ 0,\ \text{according as}\ n >4\ \text{or}\ n \leq 4$ . Note that the subspace

$$ \begin{align*} V_x:=\text {span}\,\{(L(A)+A\Phi _{\unicode{x3bb} ,f,y}-\Phi _{\unicode{x3bb} ,f,y}A)x,Ax,x\}=\text {span}\,\{(L(A)-D_{\unicode{x3bb} }(A))x,Ax,x\} \end{align*} $$

is independent of the choice of $f, y$ . Its dimension is no larger than three. Since $y+N_\alpha \cap \ker f$ is a coset of an infinite-dimensional subspace, we claim that $y+N_\alpha \cap \ker f\not \subseteq V_x$ . Pick $y_x\in (y+N_\alpha \cap \ker f)\setminus V_x$ . Then $y_x\in \mathcal {I}(f)$ . Hence, (3-29) holds true for $y_x$ and $\Phi _{\unicode{x3bb} ,f,y_x}$ . The fact that $y_x\notin V_x$ implies that the coefficient of $y_x$ in (3-29) is $0$ when we substitute $y_x$ for y. We therefore have

$$ \begin{align*} (L(A)-[\Phi_{\unicode{x3bb},f,y},A])x=(L(A)-[\Phi_{\unicode{x3bb},f,y_x},A])x\in\text{span}\,\{Ax,x\}\subseteq N_{\unicode{x3bb}}. \end{align*} $$

It follows from $y\notin N_{\unicode{x3bb} }$ that the coefficient of y in (3-29) is $0$ (otherwise $y\in \text {span}\,\{(L(A)-[\Phi _{\unicode{x3bb} ,f,y},A])x,Ax,x\} \ \subseteq N_{\unicode{x3bb} }$ , which is a contradiction). Considering the coefficients of $Ax$ and x in (3-29), we see that

$$ \begin{align*} (L(A)-D_{\unicode{x3bb}}(A))x=(L(A)-[\Phi_{\unicode{x3bb},f,y},A])x=c_{\unicode{x3bb}} Ax+H_{\unicode{x3bb}}(A)x, \end{align*} $$

where $c_{\unicode{x3bb} }\in \mathbb {C}$ merely depends on $\lambda, f$ and y, and not on x, and $H_\lambda(A) \in \mathbb{C}I$ depends only on $\lambda,a,f$ and y, and not on x.

Let us next show that $c_{\unicode{x3bb} }=0$ . Suppose that $c_{\unicode{x3bb} }\neq 0$ . For any $A_1,\ldots ,A_n\in \text {Alg}\mathcal {N}$ and an arbitrary $x\in N_{\unicode{x3bb} }$ , by the above discussion,

$$ \begin{align*} &D_{\unicode{x3bb}}(p_n(A_1,\ldots,A_n))x+c_{\unicode{x3bb}} p_n(A_1,\ldots,A_n)x+H_{\unicode{x3bb}}(p_n(A_1,\ldots,A_n))x\\[-2pt] &\quad=L(p_n(A_1,\ldots,A_n))x\\[-2pt] &\quad=\sum_{j=1}^np_n(A_1,\ldots,D_{\unicode{x3bb}}(A_j)+c_{\unicode{x3bb}} A_j,\ldots,A_n)x\\[-2pt] &\quad=D_{\unicode{x3bb}}(p_n(A_1,\ldots,A_n))x+n\cdot c_{\unicode{x3bb}} p_n(A_1,\ldots,A_n)x. \end{align*} $$

Here, the last equality is due to the fact that $D_{\unicode{x3bb} }$ is a derivation and so a Lie n-derivation (Proposition 2.4). This shows that

$$ \begin{align*} (n-1)\cdot c_{\unicode{x3bb}} p_n(A_1,\ldots,A_n)x=H_{\unicode{x3bb}}(p_n(A_1,\ldots,A_n))x. \end{align*} $$

Restricting the above equation to $N_{\unicode{x3bb} }$ , by Lemma 2.6,

$$ \begin{align*} \frac{1}{(n-1)c_{\unicode{x3bb}}}H_{\unicode{x3bb}}(p_n(A_1,\ldots,A_n))=0. \end{align*} $$

That is, $p_n(A_1,\ldots ,A_n)|_{N_{\unicode{x3bb} }}=0$ , which is absurd. In fact, for an arbitrary $N\in \mathcal {N}\setminus \{\{0\}\}$ satisfying $N\subsetneqq N_{\unicode{x3bb} }$ and $v\in N_{\unicode{x3bb} }\setminus N$ , one can define a bounded linear functional $\widetilde w_0$ on the subspace $\text {span}\,\{v+N\}$ of $\mathcal {X}/N$ by $\widetilde w_0(av+N):=a$ . By the Hahn–Banach theorem, we get an extension $\widetilde w\in (\mathcal {X}/N)^*$ . Let $w=\widetilde w\circ \theta $ , where $\theta \colon \mathcal {X}\longrightarrow \mathcal {X}/N$ is the quotient mapping. Then $w\in N^\bot $ and $w(v)=1$ . For any $u\in N\setminus \{0\}$ , we have $u\otimes w=p_n(u\otimes w,v\otimes w,\ldots , v\otimes w)=0$ , while $u\otimes w(v)=u\neq 0$ , which is a contradiction.

Let us now consider the case of $n=2$ . For any $f\in N_{\unicode{x3bb} }^\bot \setminus \{0\}$ , $y\in \mathcal {I}(f)$ and an arbitrary $x\in N_{\unicode{x3bb} }$ ,

(3-30) $$ \begin{align} L(p_2(A,x\otimes f))&=L(Ax\otimes f-x\otimes A^*f)\nonumber\\ &=\Phi_{\unicode{x3bb},f,y}Ax\otimes f+Ax\otimes \varphi_{\unicode{x3bb},f,y}+h_{\unicode{x3bb}}(Ax,f)I\nonumber\\ &\quad-\Phi_{\unicode{x3bb},f,y}x\otimes A^*f-x\otimes\psi_{\unicode{x3bb},A^*f}-h_{\unicode{x3bb}}(x,A^*f)I \end{align} $$

and

(3-31) $$ \begin{align} L(p_2(A,x\otimes f))&=p_2(L(A),x\otimes f)+p_2(A,L(x\otimes f))\nonumber\\ &=L(A)x\otimes f-x\otimes L(A)^*f+A\Phi_{\unicode{x3bb},f,y}x\otimes f+Ax\otimes \varphi_{\unicode{x3bb},f,y}\nonumber\\ &\quad-\Phi_{\unicode{x3bb},f,y}x\otimes A^*f-x\otimes A^*\varphi_{\unicode{x3bb},f,y}. \end{align} $$

Combining (3-30) with (3-31) gives

(3-32) $$ \begin{align} (L(A)+A\Phi_{\unicode{x3bb},f,y}-\Phi_{\unicode{x3bb},f,y}A)x\otimes f&=x\otimes L(A)^*f+x\otimes A^*\varphi_{\unicode{x3bb},f,y}\notag\\ &\quad+h_{\unicode{x3bb}}(Ax,f)I-x\otimes \psi_{\unicode{x3bb},A^*f}-h_{\unicode{x3bb}}(x,A^*f)I. \end{align} $$

Since I is of infinite rank,

$$ \begin{align*} h_{\unicode{x3bb}}(Ax,f)-h_{\unicode{x3bb}}(x,A^*f)=0. \end{align*} $$

Considering the action of (3-32) on y, we finish the proof of this case.

Finally (from the equality $L(\cdot )|_{N_{\unicode{x3bb} }}{{\kern-0.9pt}={\kern-0.9pt}}D_{\unicode{x3bb} }(\cdot ){{\kern-0.9pt}+{\kern-0.9pt}}H_{\unicode{x3bb} }(\cdot )),\,D_{\unicode{x3bb} }(\cdot ){{\kern-0.9pt}={\kern-0.9pt}}L(\cdot )|_{N_{\unicode{x3bb} }}{{\kern-0.9pt}-{\kern-0.9pt}}H_{\unicode{x3bb} }(\cdot )\colon \text {Alg} \mathcal {N}\longrightarrow \mathcal {B}(N_{\unicode{x3bb} },\mathcal {X})$ . If L is continuous, then $\Phi _{\unicode{x3bb} ,f,y}$ is continuous and hence $D_{\unicode{x3bb} }$ is bounded. By the relation $L(\cdot )|_{N_{\unicode{x3bb} }}=D_{\unicode{x3bb} }(\cdot )+H_{\unicode{x3bb} }(\cdot )$ , we get the continuity of $H_{\unicode{x3bb} }$ .

Proof. As we mention at the beginning of this subsection, it suffices to consider the case of $\dim X>2$ , $\mathcal {X}_-=\mathcal {X}$ , and $\{0\}=\{0\}_+$ . For arbitrary $\alpha ,\beta \in \mathit {\Lambda }$ , suppose that $N_\alpha \subseteq N_\beta $ . For any $x\in N_\alpha $ , fixing $f\in N_\beta ^\bot \setminus \{0\}$ and $y\in \mathcal {I}(f)$ , then by (3-26), we know that the equality

(3-33) $$ \begin{align} (\Phi_{\alpha,f_1,y_1}-\Phi_{\beta,f_2,y_2})x\otimes f=x\otimes(\psi_{\beta,f_2}-\psi_{\alpha,f_1})+(h_\beta(x,f)-h_\alpha(x,f))I \end{align} $$

holds for all $f_1\in N_\alpha ^\bot $ , $f_2\in N_\beta ^\bot $ , and $y_j\in \mathcal {I}(f_j)$ . Note that I has infinite rank, and thus $h_\beta (x,f)-h_\alpha (x,f)=0$ . By the action of (3-33) on y, $\Phi _{\alpha ,f_1,y_1}-\Phi _{\beta ,f_2,y_2}|_{N_\alpha }\in \mathbb {C}I|_{N_\alpha }$ . By Lemma 3.20, we conclude that for all $A\in\text{Alg}\mathcal{N}$

$$ \begin{align*} 0=[\Phi_{\alpha,f_1,y_1}-\Phi_{\beta,f_2,y_2}|_{N_\alpha},A]x=(H_\beta(A)-H_\alpha(A))x. \end{align*} $$

Hence $H_\alpha (A)=H_\beta (A)|_{N_\alpha }$ . For each $\unicode{x3bb} \in \mathit {\Lambda }$ , let us write $\widetilde H_{\unicode{x3bb} }(A)I|_{N_{\unicode{x3bb} }}=H_{\unicode{x3bb} }(A)$ . By Lemma 3.20, it follows that $\widetilde H_{\unicode{x3bb} }\colon \text {Alg}\mathcal {N}\longrightarrow \mathbb {C}I$ is a (continuous, if L is continuous) linear functional and $\widetilde H_\alpha (A)=\widetilde H_\beta (A)$ . Define $H(A):=\widetilde H_{\unicode{x3bb} }(A)I$ for some $\unicode{x3bb} \in \mathit {\Lambda }$ and write $D:=L-H$ . Thus $H\colon \text {Alg}\mathcal {N}\longrightarrow \mathbb {C}I$ is a (bounded, if L is continuous) linear operator. For any $A,B\in \text {Alg}\mathcal {N}$ and $x\in \bigcup _{\unicode{x3bb} \in \mathit {\Lambda }}N_{\unicode{x3bb} }$ , there is a $\unicode{x3bb} \in \mathit {\Lambda }$ such that $x\in N_{\unicode{x3bb} }$ . By invoking Lemma 3.20 again it follows that

$$ \begin{align*} D(AB)x&=(L-H)(AB)x\\ &=D_{\unicode{x3bb}}(AB)x\\ &=D_{\unicode{x3bb}}(A)Bx+AD_{\unicode{x3bb}}(B)x\\ &=(L(A)-\widetilde H_{\unicode{x3bb}}(A)I|_{N_{\unicode{x3bb}}})Bx+A(L(B)-\widetilde H_{\unicode{x3bb}}(B)I|_{N_{\unicode{x3bb}}})x\\ &=(L(A)-\widetilde H_{\unicode{x3bb}}(A)I)Bx+A(L(B)-\widetilde H_{\unicode{x3bb}}(B)I)x\\ &=(D(A)B+AD(B))x. \end{align*} $$

Taking into account the fact $\mathcal {X}=\overline {\text {span}\,\bigcup _{\unicode{x3bb} \in \mathit {\Lambda }}N_{\unicode{x3bb} }}=\overline {\bigcup _{\unicode{x3bb} \in \mathit {\Lambda }}N_{\unicode{x3bb} }}$ (since $\mathcal {N}$ is totally ordered, $\bigcup _{\unicode{x3bb} \in \mathit {\Lambda }}N_{\unicode{x3bb} }$ is a linear space), we know that $D\colon \text {Alg}\mathcal {N}\longrightarrow \mathcal {B(X)}$ is a (continuous, if L is continuous) derivation. By invoking Proposition 2.4, we see that H vanishes on all $(n-1)\,$ th commutators on $\text {Alg}\mathcal {N}$ .